A reverse extended Hardy–Hilbert’s inequality with parameters

Abstract

In this paper, by virtue of the symmetry principle, applying the techniques of real analysis and Euler–Maclaurin summation formula, we construct proper weight coefficients and use them to establish a reverse extended Hardy–Hilbert’s inequality with multi-parameters. Then, we obtain the equivalent forms and some equivalent statements of the best possible constant factor related to several parameters. Finally, we illustrate how the obtained results can generate some new reverse Hardy–Hilbert-type inequalities.

1 Introduction

Suppose that $$p > 1$$, $$\frac{1}{p} + \frac{1}{q} = 1$$, $$a_{m}, b_{n} \ge 0$$, $$0 < \sum_{m = 1}^{\infty} a_{m}^{p} < \infty$$, and $$0 < \sum_{n = 1}^{\infty} b_{n}^{q} < \infty$$. We have the following well-known Hardy–Hilbert’s inequality with the best possible constant factor $$\frac{\pi}{\sin (\pi /p)}$$ (cf. [1], Theorem 315):

$$\sum_{m = 1}^{\infty} \sum _{n = 1}^{\infty} \frac{a_{m}b_{n}}{m + n} < \frac{\pi}{\sin (\pi /p)}\Biggl(\sum_{m = 1}^{\infty} a_{m}^{p} \Biggr)^{\frac{1}{p}}\Biggl(\sum _{n = 1}^{\infty} b_{n}^{q} \Biggr)^{\frac{1}{q}}.$$
(1)

In 2006, by introducing parameters $$\lambda _{i} \in (0,2]$$ ($$i = 1,2$$), $$\lambda _{1} + \lambda _{2} = \lambda \in (0,4]$$, using Euler–Maclaurin summation formula, an extension of (1) was provided by Krnić et al. [2] as follows:

$$\sum_{m = 1}^{\infty} \sum _{n = 1}^{\infty} \frac{a_{m}b_{n}}{(m + n)^{\lambda}} < B(\lambda _{1},\lambda _{2})\Biggl[\sum _{m = 1}^{\infty} m^{p(1 - \lambda _{1}) - 1}a_{m}^{p} \Biggr]^{\frac{1}{p}}\Biggl[\sum_{n = 1}^{\infty} n^{q(1 - \lambda _{2}) - 1}b_{n}^{q} \Biggr]^{\frac{1}{q}},$$
(2)

where the constant factor $$B(\lambda _{1},\lambda _{2})$$ is the best possible.

$$B(u,v) = \int _{0}^{\infty} \frac{t^{u - 1}}{(1 + t)^{u + v}}\,dt\quad (u,v > 0)$$

is the beta function. For $$\lambda = 1$$, $$\lambda _{1} = \frac{1}{q}$$, $$\lambda _{2} = \frac{1}{p}$$, inequality (2) reduces to (1); for $$p = q = 2$$, $$\lambda _{1} = \lambda _{2} = \frac{\lambda}{2}$$, (2) reduces to Yang’s inequality in [3]. Recently, applying inequality (2), Adiyasuren et al. [4] gave a new Hardy–Hilbert’s inequality with the kernel $$\frac{1}{(m + n)^{\lambda}}$$ involving two partial sums.

If $$f(x),g(y) \ge 0$$, $$0 < \int _{0}^{\infty} f^{p}(x)\,dx < \infty$$, and $$0 < \int _{0}^{\infty} g^{q}(y)\,dy < \infty$$, then we still have the following Hardy–Hilbert’s integral inequality (cf. [1], Theorem 316):

$$\int _{0}^{\infty} \int _{0}^{\infty} \frac{f(x)g(y)}{x + y}\,dx\,dy < \frac{\pi}{\sin (\pi /p)}\biggl( \int _{0}^{\infty} f^{p}(x)\,dx \biggr)^{\frac{1}{p}}\biggl( \int _{0}^{\infty} g^{q}(y)\,dy \biggr)^{\frac{1}{q}},$$
(3)

where the constant factor $$\pi /\sin (\frac{\pi}{p})$$ is still the best possible. Inequalities (1), (2), and (3) with their extensions and reverses play an important role in the analysis and its applications (cf. [515]).

In 1934, a half-discrete Hilbert-type inequality was given as follows (cf. [1], Theorem 351): If $$K(t)$$ ($$t > 0$$) is a decreasing function, $$p > 1$$, $$\frac{1}{p} + \frac{1}{q} = 1$$, $$0 < \phi (s) = \int _{0}^{\infty} K(t)t^{s - 1}\,dt < \infty$$, $$a_{n} \ge 0$$, $$0 < \sum_{n = 1}^{\infty} a_{n}^{p} < \infty$$, then we have

$$\int _{0}^{\infty} x^{p - 2}\Biggl(\sum _{n = 1}^{\infty} K(nx)a_{n} \Biggr)^{p}\,dx < \phi ^{p}\biggl(\frac{1}{q} \biggr)\sum_{n = 1}^{\infty} a_{n}^{p}.$$
(4)

In recent years, some new extensions of (4) with the reverses were provided by [1623].

In 2016, by means of the technique of real analysis and the weight coefficients, Hong et al. [24] considered some equivalent statements of the extensions of (1) with the best possible constant factor related to a few parameters. Other similar works about the extensions of (1), (2), (3), and (4) with the reverses were given by [2532].

In this paper, following the way of [2, 24], by means of the weight coefficients, the idea of introduced parameters, the techniques of real analysis, and the Euler–Maclaurin summation formula, a new reverse of the extension of (1) with parameters as well as the equivalent forms are given. The equivalent statements of the best possible constant factor related to several parameters are obtained, and some particular inequalities are provided.

2 Some lemmas

In what follows, we suppose that $$0 < p < 1$$ ($$q < 0$$), $$\frac{1}{p} + \frac{1}{q} = 1$$, $$\lambda \in (0,\frac{5}{2}]$$, $$\lambda _{i} \in (0,\frac{5}{4}] \cap (0,\lambda )$$ ($$i = 1,2$$), $$a_{m}, b_{n} \ge 0$$, $$m,n \in \mathrm{N} = \{ 1,2, \ldots \}$$, such that

$$0 < \sum_{m = 1}^{\infty} m^{p[1 - (\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q})] - 1} a_{m}^{p} < \infty ,\quad\text{and}\quad0 < \sum _{n = 1}^{\infty} n^{q[1 - (\frac{\lambda _{2}}{p} + \frac{\lambda - \lambda _{1}}{q})] - 1} b_{n}^{q} < \infty .$$

Lemma 1

(cf. [5], (2.2.13))

If $$g(t)$$ is a positive strictly decreasing function in $$[m,\infty )$$ ($$m \in \mathrm{N}$$) with $$g(\infty ) = 0$$, $$P_{i}(t)$$ and $$B_{i}$$ ($$i \in \mathrm{N}$$) are the Bernoulli functions and the Bernoulli numbers of i-order, then we have

$$\int _{m}^{\infty} P_{2q - 1} (t)g(t)\,dt = \varepsilon \frac{B_{2q}}{q}\biggl(\frac{1}{2^{2q}} - 1\biggr)g(m) \quad (0 < \varepsilon < 1;q = 1,2, \ldots ).$$
(5)

In particular, for $$q = 1$$, in view of $$B_{2} = \frac{1}{6}$$, we have

$$- \frac{1}{8}g(m) < \int _{m}^{\infty} P_{1} (t)g(t)\,dt < 0.$$
(6)

Lemma 2

Define the following weight coefficient:

$$\varpi (\lambda _{2},m): = m^{\lambda - \lambda _{2}}\sum _{n = 1}^{\infty} \frac{n^{\lambda _{2} - 1}}{m^{\lambda} + n^{\lambda}}\quad (m \in \mathrm{N}).$$
(7)

We have the following inequalities:

$$\frac{\pi}{\lambda \sin (\pi \lambda _{2}/\lambda )}\bigl(1 - \theta _{m}(\lambda _{2})\bigr) < \varpi (\lambda _{2},m) < k_{\lambda} (\lambda _{2}): = \frac{\pi}{\lambda \sin (\pi \lambda _{2}/\lambda )}\quad (m \in \mathrm{N}),$$
(8)

where $$\theta _{m}(\lambda _{2})$$ is indicated by

$$\theta _{m}(\lambda _{2}): = \frac{\sin (\pi \lambda _{2}/\lambda )}{\pi} \int _{0}^{\frac{1}{m^{\lambda}}} \frac{u^{(\lambda _{2}/\lambda ) - 1}}{1 + u}\,du = O \biggl(\frac{1}{m^{\lambda _{2}}}\biggr) \in (0,1)\quad (m \in \mathrm{N}).$$
(9)

Proof

For fixed $$m \in \mathrm{N}$$, we set the following function:

$$g(m,t): = \frac{t^{\lambda _{2} - 1}}{m^{\lambda} + t^{\lambda}}\quad (t > 0).$$

By the use of Euler–Maclaurin summation formula (cf. [2, 3]), we have

\begin{aligned}& \begin{aligned} \sum_{n = 1}^{\infty} g(m,n) &= \int _{1}^{\infty} g(m,t)\,dt + \frac{1}{2} g(m,1) + \int _{1}^{\infty} P_{1}(t)g'(m,t)\,dt \\ &= \int _{0}^{\infty} g(m,t)\,dt - h(m), \end{aligned}\\& h(m): = \int _{0}^{1} g(m,t)\,dt - \frac{1}{2}g(m,1) - \int _{1}^{\infty} P_{1}(t)g'(m,t)\,dt. \end{aligned}

We find $$\frac{1}{2}g(m,1) = \frac{1}{2(m^{\lambda} + 1)}$$,

\begin{aligned} \int _{0}^{1} g(m,t)\,dt &= \int _{0}^{1} \frac{t^{\lambda _{2} - 1}}{m^{\lambda} + t^{\lambda}}\,dt = \frac{1}{\lambda _{2}} \int _{0}^{1} \frac{dt^{\lambda _{2}}}{m^{\lambda} + t^{\lambda}} \\ &= \frac{1}{\lambda _{2}}\frac{t^{\lambda _{2}}}{m^{\lambda} + t^{\lambda}} \biggm|_{0}^{1} + \frac{\lambda}{\lambda _{2}} \int _{0}^{1} \frac{t^{\lambda + \lambda _{2} - 1}}{(m^{\lambda} + t^{\lambda} )^{2}}\,dt > \frac{1}{\lambda _{2}}\frac{1}{m^{\lambda} + 1}. \end{aligned}

We also obtain

\begin{aligned} g'(m,t) &= \frac{(\lambda _{2} - 1)t^{\lambda _{2} - 2}}{m^{\lambda} + t^{\lambda}} - \frac{\lambda t^{\lambda + \lambda _{2} - 2}}{(m^{\lambda} + t^{\lambda} )^{2}} = - \frac{(1 - \lambda _{2})t^{\lambda _{2} - 2}}{m^{\lambda} + t^{\lambda}} - \frac{\lambda (m^{\lambda} + t^{\lambda} - m^{\lambda} )t^{\lambda _{2} - 2}}{(m^{\lambda} + t^{\lambda} )^{2}} \\ &= - \frac{(\lambda + 1 - \lambda _{2})t^{\lambda _{2} - 2}}{m^{\lambda} + t^{\lambda}} + \frac{\lambda m^{\lambda} t^{\lambda _{2} - 2}}{(m^{\lambda} + t^{\lambda} )^{2}}. \end{aligned}

For $$0 < \lambda _{2} \le \frac{5}{4}$$, $$\lambda _{2} < \lambda \le \frac{5}{2}$$, it follows that

$$\frac{d}{dt}\biggl[\frac{t^{\lambda _{2} - 2}}{(m^{\lambda} + t^{\lambda} )^{i}}\biggr] < 0\quad (i = 1,2).$$

By (6), we obtain

\begin{aligned}& (\lambda + 1 - \lambda _{2}) \int _{1}^{\infty} P_{1}(t) \frac{t^{\lambda _{2} - 2}}{m^{\lambda} + t^{\lambda}}\,dt > - \frac{\lambda + 1 - \lambda _{2}}{8(m^{\lambda} + 1)},\\& - m^{\lambda} \lambda \int _{1}^{\infty} P_{1}(t) \frac{t^{\lambda _{2} - 2}}{(m^{\lambda} + t^{\lambda} )^{2}}\,dt > 0, \end{aligned}

and then we find

$$- \int _{1}^{\infty} P_{1}(t)g'(m,t)\,dt > - \frac{\lambda + 1 - \lambda _{2}}{8(m^{\lambda} + 1)} + 0 = - \frac{\lambda + 1 - \lambda _{2}}{8(m^{\lambda} + 1)}.$$

Hence, it follows that

\begin{aligned} h(m)& > \frac{1}{\lambda _{2}}\frac{1}{m^{\lambda} + 1} - \frac{1}{2(m^{\lambda} + 1)} - \frac{\lambda + 1 - \lambda _{2}}{8(m^{\lambda} + 1)} = \frac{8 - (5 + \lambda )\lambda _{2} + \lambda _{2}^{2}}{8\lambda _{2}(m^{\lambda} + 1)}\\ &\ge \frac{8 - (5 + \frac{5}{2})\lambda _{2} + \lambda _{2}^{2}}{8\lambda _{2}(m^{\lambda} + 1)} = \frac{16 - 15\lambda _{2} + 2\lambda _{2}^{2}}{16\lambda _{2}(m^{\lambda} + 1)}. \end{aligned}

Since $$(6 - 15\lambda _{2} + 2\lambda _{2}^{2})' = - 15 + 4\lambda _{2} < 0$$ ($$\lambda _{2} \in (0,\frac{5}{4}]$$), we have

$$h(m) > \frac{16 - 15(\frac{5}{4}) + 2(\frac{5}{4})^{2}}{16\lambda _{2}(m^{\lambda} + 1)} = \frac{3}{128\lambda _{2}(m^{\lambda} + 1)} > 0.$$

Setting $$t = mu^{1/\lambda}$$, we find

\begin{aligned} \varpi (\lambda _{2},m) &= m^{\lambda - \lambda _{2}}\sum _{n = 1}^{\infty} g(m,n) < m^{\lambda - \lambda _{2}} \int _{0}^{\infty} g(m,t)\,dt \\ &= m^{\lambda - \lambda _{2}} \int _{0}^{\infty} \frac{t^{\lambda _{2} - 1}}{m^{\lambda} + t^{\lambda}}\,dt = \frac{1}{\lambda} \int _{0}^{\infty} \frac{u^{(\lambda _{2}/\lambda ) - 1}}{1 + u}\,du = \frac{\pi}{\lambda \sin (\pi \lambda _{2}/\lambda )}. \end{aligned}

On the other hand, we also have

\begin{aligned}& \begin{aligned} \sum_{n = 1}^{\infty} g(m,n) &= \int _{1}^{\infty} g(m,t)\,dt + \frac{1}{2} g(m,1) + \int _{1}^{\infty} P_{1}(t)g'(m,t)\,dt \\ &= \int _{1}^{\infty} g(m,t)\,dt + H(m), \end{aligned} \\& H(m): = \frac{1}{2}g(m,1) + \int _{1}^{\infty} P_{1}(t)g'(m,t)\,dt. \end{aligned}

Since we find $$\frac{1}{2}g(m,1) = \frac{1}{2(m^{\lambda} + 1)}$$ and

$$g'(m,t) = - \frac{(\lambda + 1 - \lambda _{2})t^{\lambda _{2} - 2}}{m^{\lambda} + t^{\lambda}} + \frac{\lambda m^{\lambda} t^{\lambda _{2} - 2}}{(m^{\lambda} + t^{\lambda} )^{2}},$$

in view of (6), we obtain

\begin{aligned}& - (\lambda + 1 - \lambda _{2}) \int _{1}^{\infty} P_{1}(t) \frac{t^{\lambda _{2} - 2}}{m^{\lambda} + t^{\lambda}}\,dt > 0,\quad \text{and} \\& \lambda m^{\lambda} \int _{1}^{\infty} P_{1}(t) \frac{t^{\lambda _{2} - 2}}{(m^{\lambda} + t^{\lambda} )^{2}}\,dt > - \frac{\lambda m^{\lambda}}{ 8(m{}^{\lambda} + 1)^{2}}. \end{aligned}

Hence, we have

$$H(m) > \frac{1}{2(m^{\lambda} + 1)} - \frac{\lambda m^{\lambda}}{ 8(m^{\lambda} + 1)^{2}} > \frac{4}{8(m^{\lambda} + 1)} - \frac{5/2}{8(m^{\lambda} + 1)} > 0.$$

Setting $$t = mu^{1/\lambda}$$, we obtain

\begin{aligned}& \begin{aligned} \varpi (\lambda _{2},m)& = m^{\lambda - \lambda _{2}}\sum _{n = 1}^{\infty} g(m,n) > m^{\lambda - \lambda _{2}} \int _{1}^{\infty} g(m,t)\,dt \\ &= m^{\lambda - \lambda _{2}} \int _{0}^{\infty} g(m,t)\,dt - m^{\lambda - \lambda _{2}} \int _{0}^{1} g(m,t)\,dt \\ &= \frac{\pi}{\lambda \sin (\pi \lambda _{2}/\lambda )}\biggl[1 - \frac{\lambda \sin (\pi \lambda _{2}/\lambda )}{\pi} m^{\lambda - \lambda _{2}} \int _{0}^{1} \frac{t^{\lambda _{2} - 1}}{m^{\lambda} + t^{\lambda}}\,dt \biggr] \\ &= \frac{\pi}{\lambda \sin (\pi \lambda _{2}/\lambda )}\bigl(1 - \theta _{m}(\lambda _{2})\bigr) > 0,\end{aligned} \end{aligned}

where $$\theta _{m}(\lambda _{2}) = \frac{\sin (\pi \lambda _{2}/\lambda )}{\pi} \int _{0}^{\frac{1}{m^{\lambda}}} \frac{u^{(\lambda _{2}/\lambda ) - 1}}{1 + u}\,du$$. Since we find

$$0 < \int _{0}^{\frac{1}{m^{\lambda}}} \frac{u^{(\lambda _{2}/\lambda ) - 1}}{1 + u}\,du < \int _{0}^{\frac{1}{m^{\lambda}}} u^{(\lambda _{2}/\lambda ) - 1}\,du = \frac{\lambda}{\lambda _{2}m^{\lambda _{2}}},$$

namely, $$\theta _{m}(\lambda _{2}) = O(\frac{1}{m^{\lambda _{2}}}) \in (0,1)$$ ($$m \in \mathrm{N}$$). Therefore, inequalities (8) with (9) follow.

The lemma is proved. □

Lemma 3

We have the following reverse extended Hardy–Hilbert’s inequality with parameters:

\begin{aligned} I ={}& \sum_{n = 1}^{\infty} \sum _{m = 1}^{\infty} \frac{a_{m}b_{n}}{m^{\lambda} + n^{\lambda}} > k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}( \lambda _{1}) \\ &{}\times \Biggl\{ \sum_{m = 1}^{\infty} \bigl(1 - \theta _{m}(\lambda _{2})\bigr)m^{p[1 - (\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q})] - 1} a_{m}^{p}\Biggr\} ^{\frac{1}{p}}\Biggl\{ \sum _{n = 1}^{\infty} n^{q[1 - (\frac{\lambda _{2}}{p} + \frac{\lambda - \lambda _{1}}{q})] - 1} b_{n}^{q}\Biggr\} ^{\frac{1}{q}}. \end{aligned}
(10)

Proof

In the same way, for $$n \in \mathbf{N}$$, we have the following inequalities for another weight coefficient:

\begin{aligned}& \begin{gathered} \omega (\lambda _{1},n): = n^{\lambda - \lambda _{1}}\sum _{m = 1}^{\infty} \frac{m^{\lambda _{1} - 1}}{m^{\lambda} + n^{\lambda}} \quad (n \in \mathrm{N}),\\ \frac{\pi}{\lambda \sin (\pi \lambda _{1}/\lambda )}\bigl(1 - \theta _{n}(\lambda _{1})\bigr) < \omega (\lambda _{1},n) < k_{\lambda} (\lambda _{1}) = \frac{\pi}{\lambda \sin (\pi \lambda _{1}/\lambda )}, \end{gathered} \end{aligned}
(11)
\begin{aligned}& \theta _{n}(\lambda _{1}) = \frac{\sin (\pi \lambda _{1}/\lambda )}{\pi} \int _{0}^{\frac{1}{n^{\lambda}}} \frac{u^{(\lambda _{1}/\lambda ) - 1}}{1 + u}\,du = O \biggl(\frac{1}{n^{\lambda _{1}}}\biggr) \in (0,1)\quad (n \in \mathrm{N}). \end{aligned}
(12)

By the reverse Hölder inequality (cf. [33]), we obtain

\begin{aligned} I& = \sum_{n = 1}^{\infty} \sum_{m = 1}^{\infty} \frac{1}{m^{\lambda} + n^{\lambda}} \biggl[\frac{n^{(\lambda _{2} - 1)/p}}{m^{(\lambda _{1} - 1)/q}}a_{m}\biggr] \biggl[ \frac{m^{(\lambda _{1} - 1)/q}}{n^{(\lambda _{2} - 1)/p}}b_{n}\biggr] \\ &\ge \Biggl[\sum_{m = 1}^{\infty} \sum _{n = 1}^{\infty} \frac{1}{m^{\lambda} + n^{\lambda}} \frac{n^{\lambda _{2} - 1}}{m^{(\lambda _{1} - 1)(p - 1)}}a_{m}^{p}\Biggr]^{\frac{1}{p}} \Biggl[\sum_{n = 1}^{\infty} \sum _{m = 1}^{\infty} \frac{1}{m^{\lambda} + n^{\lambda}} \frac{m^{\lambda _{1} - 1}}{n^{(\lambda _{2} - 1)(q - 1)}}b_{n}^{q}\Biggr]^{\frac{1}{q}} \\ &= \Biggl\{ \sum_{m = 1}^{\infty} \varpi ( \lambda _{2},m) m^{p[1 - (\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q})] - 1}a_{m}^{p} \Biggr\} ^{\frac{1}{p}}\Biggl\{ \sum_{n = 1}^{\infty} \omega (\lambda _{1},n) n^{q[1 - (\frac{\lambda _{2}}{p} + \frac{\lambda - \lambda _{1}}{q})] - 1}b_{n}^{q} \Biggr\} ^{\frac{1}{q}}. \end{aligned}

Then, by (8) and (11), in view of $$0 < p < 1$$ ($$q < 0$$), we have (10).

The lemma is proved. □

Remark 1

By (10), for $$\lambda _{1} + \lambda _{2} = \lambda \in (0,\frac{5}{2}]$$, $$0 < \lambda _{i} \le \frac{5}{4}$$ ($$i = 1,2$$), we find

$$0 < \sum_{m = 1}^{\infty} m^{p(1 - \lambda _{1}) - 1} a_{m}^{p} < \infty ,\qquad 0 < \sum _{n = 1}^{\infty} n^{q(1 - \lambda _{2}) - 1} b_{n}^{q} < \infty$$

and the following reverse inequality:

$$\sum_{n = 1}^{\infty} \sum _{m = 1}^{\infty} \frac{a_{m}b_{n}}{m^{\lambda} + n^{\lambda}} > k_{\lambda} (\lambda _{1}) \Biggl[\sum _{m = 1}^{\infty} \bigl(1 - \theta _{m}( \lambda _{2})\bigr)m^{p(1 - \lambda _{1}) - 1} a_{m}^{p} \Biggr]^{\frac{1}{p}}\Biggl[\sum_{n = 1}^{\infty} n^{q(1 - \lambda _{2}) - 1} b_{n}^{q}\Biggr]^{\frac{1}{q}}.$$
(13)

Lemma 4

The constant factor $$k_{\lambda} (\lambda _{1}) = \frac{\pi}{\lambda \sin (\pi \lambda _{1}/\lambda )}$$ in (13) is the best possible.

Proof

For any $$0 < \varepsilon < p\lambda _{1}$$, we set

$$\tilde{a}_{m}: = m^{\lambda _{1} - \frac{\varepsilon}{p} - 1},\qquad \tilde{b}_{n}: = n^{\lambda _{2} - \frac{\varepsilon}{q} - 1}\quad (m,n \in \mathrm{N}).$$

If there exists a constant $$M \ge k_{\lambda} (\lambda _{1})$$ such that (13) is valid when we replace $$k_{\lambda} (\lambda _{1})$$ by M, then in particular, by substitution of $$a_{m} = \tilde{a}_{m}$$ and $$b_{n} = \tilde{b}_{n}$$ in (13), we have

\begin{aligned}[b] \tilde{I}&: = \sum_{n = 1}^{\infty} \sum _{m = 1}^{\infty} \frac{\tilde{a}_{m}\tilde{b}_{n}}{m^{\lambda} + n^{\lambda}} \\ &> M \Biggl[\sum_{m = 1}^{\infty} \biggl(1 - O \biggl(\frac{1}{m^{\lambda _{2}}}\biggr)\biggr)m^{p(1 - \lambda _{1}) - 1} \tilde{a}_{m}^{p} \Biggr]^{\frac{1}{p}}\Biggl[\sum_{n = 1}^{\infty} n^{q(1 - \lambda _{2}) - 1} \tilde{b}_{n}^{q} \Biggr]^{\frac{1}{q}}. \end{aligned}
(14)

By (14) and the decreasingness property of series, for $$0 < p < 1$$, $$q < 0$$, we obtain

\begin{aligned} \tilde{I}& > M\Biggl[\sum_{m = 1}^{\infty} \biggl(1 - O\biggl(\frac{1}{m^{\lambda _{2}}}\biggr)\biggr)m^{p(1 - \lambda _{1}) - 1} m^{p\lambda _{1} - \varepsilon - p}\Biggr]^{\frac{1}{p}}\Biggl[\sum _{n = 1}^{\infty} n^{q(1 - \lambda _{2}) - 1} n^{q\lambda _{2} - \varepsilon - q} \Biggr]^{\frac{1}{q}} \\ &= M\Biggl[\sum_{m = 1}^{\infty} \biggl(1 - O \biggl(\frac{1}{m^{\lambda _{2}}}\biggr)\biggr)m^{ - \varepsilon - 1} \Biggr]^{\frac{1}{p}} \Biggl(\sum_{n = 1}^{\infty} n^{ - \varepsilon - 1} \Biggr)^{\frac{1}{q}} \\ &= M\Biggl(\sum_{m = 1}^{\infty} m^{ - \varepsilon - 1} - \sum_{m = 1}^{\infty} O \biggl(\frac{1}{m^{\lambda _{2} + \varepsilon + 1}}\biggr) \Biggr)^{\frac{1}{p}}\Biggl(1 + \sum _{n = 2}^{\infty} n^{ - \varepsilon - 1} \Biggr)^{\frac{1}{q}} \\ &> M\biggl( \int _{1}^{\infty} x^{ - \varepsilon - 1}\,dx - O(1) \biggr)^{\frac{1}{p}}\biggl(1 + \int _{1}^{\infty} y^{ - \varepsilon - 1}\,dy \biggr)^{\frac{1}{q}} \\ &= \frac{M}{\varepsilon} \bigl(1 - \varepsilon O(1)\bigr)^{\frac{1}{p}}( \varepsilon + 1)^{\frac{1}{q}}. \end{aligned}

By (11) and (12), setting $$\hat{\lambda}_{1} = \lambda _{1} - \frac{\varepsilon}{p} \in (0,\frac{5}{4}) \cap (0,\lambda )$$ ($$0 < \hat{\lambda}_{2} = \lambda _{2} + \frac{\varepsilon}{p} = \lambda - \hat{\lambda}_{1} < \lambda$$), we find

\begin{aligned}& \begin{aligned} \tilde{I} &= \sum_{n = 1}^{\infty} \Biggl[n^{(\lambda _{2} + \frac{\varepsilon}{p})}\sum_{m = 1}^{\infty} \frac{1}{m^{\lambda} + n^{\lambda}} m^{(\lambda _{1} - \frac{\varepsilon}{p}) - 1}\Biggr]n^{ - \varepsilon - 1} \\ &= \sum_{n = 1}^{\infty} \omega (\hat{ \lambda}_{1},n)n^{ - \varepsilon - 1} < k_{\lambda} (\hat{ \lambda}_{1}) \Biggl(1 + \sum_{n = 2}^{\infty} n^{ - \varepsilon - 1} \Biggr) \\ &< k_{\lambda} (\hat{\lambda}_{1}) \biggl(1 + \int _{1}^{\infty} x^{ - \varepsilon - 1}\,dx\biggr) = \frac{1}{\varepsilon} k_{\lambda} (\hat{\lambda}_{1}) ( \varepsilon + 1). \end{aligned} \end{aligned}

Then we have

$$k_{\lambda} \biggl(\lambda _{1} - \frac{\varepsilon}{p} \biggr) (\varepsilon + 1) > \varepsilon \tilde{I} > M\bigl(1 - \varepsilon O(1) \bigr)^{\frac{1}{p}}(\varepsilon + 1)^{\frac{1}{q}}.$$

For $$\varepsilon \to 0^{ +}$$, we find $$k_{\lambda} (\lambda _{1}) \ge M$$. Hence, $$M = k_{\lambda} (\lambda _{1})$$ is the best possible constant factor of (13). The lemma is proved. □

Remark 2

Setting $$\tilde{\lambda}_{1}: = \frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q}$$, $$\tilde{\lambda}_{2}: = \frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p}$$, we find

$$\tilde{\lambda}_{1} + \tilde{\lambda}_{2} = \frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q} + \frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p} = \frac{\lambda}{p} + \frac{\lambda}{q} = \lambda .$$

If we add the condition that $$\lambda - \lambda _{1} - \lambda _{2} \in ( - p\lambda _{1},p(\lambda - \lambda _{1}))$$,then we can find

$$0 < \tilde{\lambda}_{1} = \frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q} < \lambda ,\qquad 0 < \tilde{\lambda}_{2} = \lambda - \tilde{\lambda}_{1} < \lambda ;$$

if we add the condition that $$\lambda - \lambda _{1} - \lambda _{2} \in [p(\lambda - \lambda _{1} - \frac{5}{4},p(\frac{5}{4} - \lambda _{1})]$$,then we have $$\tilde{\lambda}_{1},\tilde{\lambda}_{2} \le \frac{5}{4}$$.

Then, with regard to the above assumptions, we can rewrite (13) as follows:

\begin{aligned} I &= \sum_{n = 1}^{\infty} \sum _{m = 1}^{\infty} \frac{a_{m}b_{n}}{m^{\lambda} + n^{\lambda}} \\ & > k_{\lambda} (\tilde{\lambda}_{1})\Biggl[\sum_{m = 1}^{\infty} \bigl(1 - \theta _{m}(\tilde{\lambda}_{2})\bigr)m^{p(1 - \tilde{\lambda}_{1}) - 1} a_{m}^{p}\Biggr]^{\frac{1}{p}}\Biggl[\sum _{n = 1}^{\infty} n^{q(1 - \tilde{\lambda}_{2}) - 1} b_{n}^{q}\Biggr]^{\frac{1}{q}}. \end{aligned}
(15)

Lemma 5

If the constant factor $$k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}(\lambda _{1})$$ in (10) is the best possible, then for

$$\lambda - \lambda _{1} - \lambda _{2} \in \bigl( - p \lambda _{1},p(\lambda - \lambda _{1})\bigr)\cap \biggl[p(\lambda - \lambda _{1} - \frac{5}{4},p\biggl( \frac{5}{4} - \lambda _{1}\biggr)\biggr]\bigl( \supset \{ 0 \} \bigr),$$

we have $$\lambda _{1} + \lambda _{2} = \lambda$$.

Proof

For $$0 < \tilde{\lambda}_{1} < \lambda$$,we have $$k_{\lambda} (\tilde{\lambda}_{1}) = \frac{\pi}{\lambda \sin (\pi \tilde{\lambda}_{1}/\lambda )} \in \mathrm{R}_{ +} = (0,\infty )$$.

If the constant factor $$k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}(\lambda _{1})$$ in (10) is the best possible, then for $$\tilde{\lambda}_{1} = \frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q}$$, $$\tilde{\lambda}_{2} = \frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p}$$, in view of the assumption and (15), we have the following inequality:

$$k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}( \lambda _{1}) \ge k_{\lambda} (\tilde{\lambda}_{1}).$$

By the reverse Hölder inequality with weight (cf. [33]), we find

\begin{aligned}& \begin{aligned}[b] k_{\lambda} (\tilde{\lambda}_{1}) &= k_{\lambda} \biggl(\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q}\biggr) \\ &= \int _{0}^{\infty} \frac{1}{1 + u^{\lambda}} u^{\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q} - 1}\,du = \int _{0}^{\infty} \frac{1}{1 + u^{\lambda}} \bigl(u^{\frac{\lambda - \lambda _{2} - 1}{p}}\bigr) \bigl(u^{\frac{\lambda _{1} - 1}{q}}\bigr)\,du \\ &\ge \biggl( \int _{0}^{\infty} \frac{1}{1 + u^{\lambda}} u^{\lambda - \lambda _{2} - 1}\,du\biggr)^{\frac{1}{p}}\biggl( \int _{0}^{\infty} \frac{1}{1 + u^{\lambda}} u^{\lambda _{1} - 1}\,du\biggr)^{\frac{1}{q}} \\ &= \biggl( \int _{0}^{\infty} \frac{1}{1 + v^{\lambda}} v^{\lambda _{2} - 1}\,dv\biggr)^{\frac{1}{p}}\biggl( \int _{0}^{\infty} \frac{1}{1 + u^{\lambda}} u^{\lambda _{1} - 1}\,du\biggr)^{\frac{1}{q}} \\ &= k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}( \lambda _{1}). \end{aligned} \end{aligned}
(16)

Hence, we find $$k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}(\lambda _{1}) = k_{\lambda} (\tilde{\lambda}_{1})$$, namely, (16) keeps the form of equality.

We observe that (16) keeps the form of equality if and only if there exist constants A and B such that they are not all zero and (cf. [33])

$$Au^{\lambda - \lambda _{2} - 1} = Bu^{\lambda _{1} - 1}\quad a.e. \text{ in } \mathrm{R}_{ +}.$$

Assuming that $$A \ne 0$$, we have $$u^{\lambda - \lambda _{2} - \lambda _{1}} = \frac{B}{A}$$ a.e. in $$\mathrm{R}_{ +}$$, and then $$\lambda - \lambda _{2} - \lambda _{1} = 0$$, namely, $$\lambda _{1} + \lambda _{2} = \lambda$$.

The lemma is proved. □

3 Main results

Theorem 1

Inequality (10) is equivalent to the following inequalities:

\begin{aligned}& \begin{aligned}[b] J&: = \Biggl[\sum_{n = 1}^{\infty} n^{p(\frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p}) - 1}\Biggl(\sum_{m = 1}^{\infty} \frac{1}{m^{\lambda} + n^{\lambda}} a_{m} \Biggr)^{p} \Biggr]^{\frac{1}{p}}\\ &> k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}( \lambda _{1})\Biggl\{ \sum_{m = 1}^{\infty} \bigl(1 - \theta _{m}(\lambda _{2}) \bigr)m^{p[1 - (\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q})] - 1} a_{m}^{p}\Biggr\} ^{\frac{1}{p}}, \end{aligned} \end{aligned}
(17)
\begin{aligned}& \begin{aligned}[b] J_{1}&: = \Biggl[\sum_{m = 1}^{\infty} \frac{m^{q(\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q}) - 1}}{(1 - \theta _{m}(\lambda _{2}))^{q - 1}}\Biggl(\sum_{n = 1}^{\infty} \frac{1}{m^{\lambda} + n^{\lambda}} b_{n} \Biggr)^{q} \Biggr]^{\frac{1}{q}}\\ &> k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}( \lambda _{1})\Biggl\{ \sum_{n = 1}^{\infty} n^{q[1 - (\frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p})] - 1} b_{n}^{q}\Biggr\} ^{\frac{1}{q}}. \end{aligned} \end{aligned}
(18)

If the constant factor in (10) is the best possible, then so is the constant factor in (17) and (18).

Proof

Suppose that (17) is valid. By the reverse Hölder inequality (cf. [33]), we have

\begin{aligned} I &= \sum_{n = 1}^{\infty} \Biggl[n^{\frac{ - 1}{p} + (\frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p})}\sum_{m = 1}^{\infty} \frac{1}{m^{\lambda} + n^{\lambda}} a_{m} \Biggr] \bigl[n^{\frac{1}{p} - (\frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p})}b_{n} \bigr] \\ &\ge J\Biggl\{ \sum_{n = 1}^{\infty} n^{q[1 - (\frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p})] - 1} b_{n}^{q}\Biggr\} ^{\frac{1}{q}}. \end{aligned}
(19)

Then by (17) we obtain (10).

On the other hand, assuming that (10) is valid, we set

$$b_{n}: = n^{p(\frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p}) - 1}\Biggl(\sum _{m = 1}^{\infty} \frac{1}{m^{\lambda} + n^{\lambda}} a_{m} \Biggr)^{p - 1},\quad n \in \mathbf{N}.$$

If $$J = \infty$$, then (17) is naturally valid; if $$J = 0$$, then it is impossible to make (17) valid, namely, $$J > 0$$. Suppose that $$0 < J < \infty$$. By (10), we have

\begin{aligned}& \sum_{n = 1}^{\infty} n^{q[1 - (\frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p})] - 1} b_{n}^{q}\\& \quad = J^{p} = I > k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}( \lambda _{1})\\& \qquad {}\times \Biggl\{ \sum_{m = 1}^{\infty} \bigl(1 - \theta _{m}(\lambda _{2})\bigr)m^{p[1 - (\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q})] - 1} a_{m}^{p}\Biggr\} ^{\frac{1}{p}} \Biggl\{ \sum _{n = 1}^{\infty} n^{q[1 - (\frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p})] - 1} b_{n}^{q}\Biggr\} ^{\frac{1}{q}},\\& \begin{aligned} J &= \Biggl\{ \sum_{n = 1}^{\infty} n^{q[1 - (\frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p})] - 1} b_{n}^{q}\Biggr\} ^{\frac{1}{p}}\\ &> k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}( \lambda _{1})\Biggl\{ \sum_{m = 1}^{\infty} \bigl(1 - \theta _{m}(\lambda _{2}) \bigr)m^{p[1 - (\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q})] - 1} a_{m}^{p}\Biggr\} ^{\frac{1}{p}}, \end{aligned} \end{aligned}

namely, (17) follows, which is equivalent to (10).

Suppose that (18) is valid. By the reverse Hölder inequality, we have

\begin{aligned} I &= \sum_{m = 1}^{\infty} \bigl[\bigl(1 - \theta _{m}(\lambda _{2})\bigr)^{\frac{1}{p}}m^{\frac{1}{q} - (\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q})}a_{m} \bigr] \Biggl[\frac{m^{\frac{ - 1}{q} + (\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q})}}{(1 - \theta _{m}(\lambda _{2}))^{1/p}}\sum_{n = 1}^{\infty} \frac{1}{m^{\lambda} + n{}^{\lambda}} b_{n} \Biggr] \\ &\ge \Biggl\{ \sum_{m = 1}^{\infty} \bigl(1 - \theta _{m}(\lambda _{2})\bigr)m^{p[1 - (\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q})] - 1} a_{m}^{p}\Biggr\} ^{\frac{1}{p}}J_{1}. \end{aligned}
(20)

Then by (18) we obtain (10). On the other hand, assuming that (10) is valid, we set

$$a_{m}: = \frac{m^{q(\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q}) - 1}}{(1 - \theta _{m}(\lambda _{2}))^{q - 1}}\Biggl(\sum _{n = 1}^{\infty} \frac{1}{m^{\lambda} + n^{\lambda}} b_{n} \Biggr)^{q - 1},\quad m \in \mathrm{N}.$$

If $$J_{1} = \infty$$, then (18) is naturally valid; if $$J_{1} = 0$$, then it is impossible to make (18) valid, namely, $$J_{1} > 0$$. Suppose that $$0 < J_{1} < \infty$$. By (10), we have

\begin{aligned}& \sum_{m = 1}^{\infty} \bigl(1 - \theta _{m}(\lambda _{2})\bigr)m^{p[1 - (\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q})] - 1} a_{m}^{p}\\& \quad = J_{1}^{q} = I > k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}( \lambda _{1})\\& \qquad {}\times \Biggl\{ \sum_{m = 1}^{\infty} \bigl(1 - \theta _{m}(\lambda _{2})\bigr)m^{p[1 - (\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q})] - 1} a_{m}^{p}\Biggr\} ^{\frac{1}{p}} \Biggl\{ \sum _{n = 1}^{\infty} n^{q[1 - (\frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p})] - 1} b_{n}^{q}\Biggr\} ^{\frac{1}{q}},\\& \begin{aligned} J_{1} &= \Biggl\{ \sum_{m = 1}^{\infty} \bigl(1 - \theta _{m}(\lambda _{2}) \bigr)m^{p[1 - (\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q})] - 1} a_{m}^{p}\Biggr\} ^{\frac{1}{q}}\\ &> k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}( \lambda _{1})\Biggl\{ \sum_{n = 1}^{\infty} n^{q[1 - (\frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p})] - 1} b_{n}^{q}\Biggr\} ^{\frac{1}{q}}, \end{aligned} \end{aligned}

namely, (18) follows, which is equivalent to (10).

Hence, inequalities (10), (17), and (18) are equivalent.

If the constant factor in (10) is the best possible, then so is the constant factor in (17) and (18). Otherwise, by (19) (or (20)), we would reach a contradiction that the constant factor in (10) is not the best possible.

The theorem is proved. □

Theorem 2

The following statements (i), (ii), (iii), and (iv) are equivalent:

1. (i)

Both $$k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}(\lambda _{1})$$ and $$k_{\lambda} (\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q})$$ are independent of p, q;

2. (ii)

$$k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}( \lambda _{1}) = k_{\lambda} (\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q})$$;

3. (iii)

$$k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}(\lambda _{1})$$ in (10) is the best possible constant factor;

4. (iv)

If $$\lambda - \lambda _{1} - \lambda _{2} \in ( - p\lambda _{1},p(\lambda - \lambda _{1}))\cap [p(\lambda - \lambda _{1} - \frac{5}{4},p(\frac{5}{4} - \lambda _{1} )]$$, then $$\lambda _{1} + \lambda _{2} = \lambda$$.

If the statement (iv) follows, namely, $$\lambda _{1} + \lambda _{2} = \lambda$$, then we have (13) and the following equivalent inequalities with the best possible constant factor $$k_{\lambda} (\lambda _{1})$$:

\begin{aligned}& \Biggl[\sum_{n = 1}^{\infty} n^{p\lambda _{2} - 1}\Biggl(\sum_{m = 1}^{\infty} \frac{1}{m^{\lambda} + n^{\lambda}} a_{m} \Biggr)^{p} \Biggr]^{\frac{1}{p}}> k_{\lambda} (\lambda _{1})\Biggl[\sum _{m = 1}^{\infty} \bigl(1 - \theta _{m}( \lambda _{2})\bigr)m^{p(1 - \lambda _{1}) - 1} a_{m}^{p} \Biggr]^{\frac{1}{p}}, \end{aligned}
(21)
\begin{aligned}& \Biggl[\sum_{m = 1}^{\infty} \frac{m^{q\lambda _{1} - 1}}{(1 - \theta _{m}(\lambda _{2}))^{q - 1}}\Biggl(\sum_{n = 1}^{\infty} \frac{1}{m^{\lambda} + n^{\lambda}} b_{n} \Biggr)^{q} \Biggr]^{\frac{1}{q}} > k_{\lambda} (\lambda _{1})\Biggl[\sum _{n = 1}^{\infty} n^{q(1 - \lambda _{2}) - 1} b_{n}^{q}\Biggr]^{\frac{1}{q}}. \end{aligned}
(22)

Proof

(i) (ii). By (i), since $$\tilde{\lambda}_{1} + \tilde{\lambda}_{2} = \lambda$$, we have

\begin{aligned}& k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}( \lambda _{1}) = \lim_{p \to 1^{ -}} \lim _{q \to - \infty} k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}(\lambda _{1}) = k_{\lambda} (\lambda _{2}),\\& k_{\lambda} \biggl(\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q}\biggr) = k_{\lambda} \biggl(\frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p}\biggr) = \lim_{p \to 1^{ -}} \lim_{q \to - \infty} k_{\lambda} \biggl(\frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p}\biggr) = k_{\lambda} ( \lambda _{2}), \end{aligned}

namely, $$k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}(\lambda _{1}) = k_{\lambda} (\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q})$$.

(ii) (iv). If $$k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}(\lambda _{1}) = k_{\lambda} (\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q})$$, then (16) keeps the form of equality. In view of the proof of Lemma 5, it follows that $$\lambda _{1} + \lambda _{2} = \lambda$$.

(iv) (i). If $$\lambda _{1} + \lambda _{2} = \lambda$$, then

$$k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}( \lambda _{1}) = k_{\lambda} \biggl(\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q}\biggr) = k_{\lambda} (\lambda _{1}),$$

which is independent of p, q. Hence, it follows that (i) (ii) (iv).

(iii) (iv). By the assumption and Lemma 5, we have $$\lambda _{1} + \lambda _{2} = \lambda$$.

(iv) (iii). By Lemma 4, for $$\lambda _{1} + \lambda _{2} = \lambda$$, $$k_{\lambda}^{\frac{1}{p}}(\lambda _{2})k_{\lambda}^{\frac{1}{q}}(\lambda _{1})( = k_{\lambda} (\lambda _{1}))$$ is the best possible constant factor of (10). Therefore, we have (iii) (iv).

Hence, the statements (i), (ii), (iii), and (iv) are equivalent.

The theorem is proved. □

Remark 3

(i) For $$\lambda _{1} = \lambda _{2} = \frac{\lambda}{2} \in (0,\frac{5}{4}]$$ ($$0 < \lambda \le \frac{5}{2}$$) in (13), (21), and (22), we have the following equivalent inequalities with the best possible constant factor $$\frac{\pi}{\lambda}$$:

\begin{aligned}& \sum_{n = 1}^{\infty} \sum _{m = 1}^{\infty} \frac{a_{m}b_{n}}{m^{\lambda} + n^{\lambda}} > \frac{\pi}{\lambda} \Biggl[\sum_{m = 1}^{\infty} \biggl(1 - \theta _{m}\biggl(\frac{\lambda}{2}\biggr) \biggr)m^{p(1 - \frac{\lambda}{2}) - 1} a_{m}^{p}\Biggr]^{\frac{1}{p}} \Biggl[\sum_{n = 1}^{\infty} n^{q(1 - \frac{\lambda}{2}) - 1} b_{n}^{q}\Biggr]^{\frac{1}{q}}, \end{aligned}
(23)
\begin{aligned}& \Biggl[\sum_{n = 1}^{\infty} n^{\frac{p\lambda}{2} - 1}\Biggl(\sum_{m = 1}^{\infty} \frac{1}{m^{\lambda} + n^{\lambda}} a_{m} \Biggr)^{p} \Biggr]^{\frac{1}{p}} > \frac{\pi}{\lambda} \Biggl\{ \sum _{m = 1}^{\infty} \biggl(1 - \theta _{m} \biggl(\frac{\lambda}{2}\biggr)\biggr)m^{p(1 - \frac{\lambda}{2}) - 1} a_{m}^{p}\Biggr]^{\frac{1}{p}}, \end{aligned}
(24)
\begin{aligned}& \Biggl[\sum_{m = 1}^{\infty} \frac{m^{\frac{q\lambda}{2} - 1}}{(1 - \theta _{m}(\frac{\lambda}{2}))^{q - 1}}\Biggl(\sum_{n = 1}^{\infty} \frac{1}{m^{\lambda} + n^{\lambda}} b_{n} \Biggr)^{q}\Biggr]^{\frac{1}{q}} > \frac{\pi}{ \lambda} \Biggl[\sum_{n = 1}^{\infty} n^{q(1 - \frac{\lambda}{2}) - 1} b_{n}^{q}\Biggr]^{\frac{1}{q}}. \end{aligned}
(25)

In particular, for $$\lambda = \frac{5}{2}$$,we have the following equivalent inequalities:

\begin{aligned}& \sum_{n = 1}^{\infty} \sum _{m = 1}^{\infty} \frac{a_{m}b_{n}}{m^{5/2} + n^{5/2}} > \frac{2\pi}{5} \Biggl[\sum_{m = 1}^{\infty} \biggl(1 - \theta _{m}\biggl(\frac{5}{4}\biggr)\biggr) \frac{a_{m}^{p}}{m^{1 + p/4}} \Biggr]^{\frac{1}{p}}\Biggl(\sum _{n = 1}^{\infty} \frac{b_{n}^{q}}{n^{1 + q/4}} \Biggr)^{\frac{1}{q}}, \end{aligned}
(26)
\begin{aligned}& \Biggl[\sum_{n = 1}^{\infty} n^{\frac{5p}{4} - 1}\Biggl(\sum_{m = 1}^{\infty} \frac{1}{m^{5/2} + n^{5/2}}a_{m} \Biggr)^{p} \Biggr]^{\frac{1}{p}} > \frac{2\pi}{ 5}\Biggl[\sum _{m = 1}^{\infty} \biggl(1 - \theta \biggl( \frac{5}{4}\biggr)\biggr)\frac{a_{m}^{p}}{m^{1 + p/4}} \Biggr]^{\frac{1}{p}}, \end{aligned}
(27)
\begin{aligned}& \Biggl[\sum_{m = 1}^{\infty} \frac{m^{\frac{5q}{4} - 1}}{(1 - \theta _{m}(\frac{5}{4}))^{q - 1}}\Biggl(\sum_{n = 1}^{\infty} \frac{1}{m^{5/2} + n^{5/2}}b_{n} \Biggr)^{q} \Biggr]^{\frac{1}{q}} > \frac{2\pi}{5}\Biggl(\sum _{n = 1}^{\infty} \frac{b_{n}^{q}}{n^{1 + q/4}} \Biggr)^{\frac{1}{q}}. \end{aligned}
(28)

(ii) For $$\lambda = 1$$, $$\lambda _{1} = \frac{1}{r}$$, $$\lambda _{2} = \frac{1}{s}$$ ($$r > 1,\frac{1}{r} + \frac{1}{s} = 1$$) in (13), (21), and (22), we have the following equivalent inequalities with the best possible constant factor $$\frac{\pi}{\sin (\pi /r)}$$:

\begin{aligned}& \sum_{n = 1}^{\infty} \sum _{m = 1}^{\infty} \frac{a_{m}b_{n}}{m + n} > \frac{\pi}{\sin (\pi /r)} \Biggl[\sum_{m = 1}^{\infty} \biggl(1 - \hat{\theta}_{m}\biggl(\frac{1}{r}\biggr) \biggr)m^{\frac{p}{s} - 1} a_{m}^{p} \Biggr]^{\frac{1}{p}}\Biggl(\sum_{n = 1}^{\infty} n^{\frac{q}{r} - 1} b_{n}^{q}\Biggr)^{\frac{1}{q}}, \end{aligned}
(29)
\begin{aligned}& \Biggl[\sum_{n = 1}^{\infty} n^{\frac{p}{s} - 1}\Biggl(\sum_{m = 1}^{\infty} \frac{1}{m + n}a_{m} \Biggr)^{p} \Biggr]^{\frac{1}{p}} > \frac{\pi}{\sin (\pi /r)}\Biggl[\sum _{m = 1}^{\infty} \biggl(1 - \hat{\theta}_{m} \biggl(\frac{1}{s}\biggr)\biggr)m^{\frac{p}{s} - 1} a_{m}^{p} \Biggr]^{\frac{1}{p}}, \end{aligned}
(30)
\begin{aligned}& \Biggl[\sum_{m = 1}^{\infty} \frac{m^{\frac{q}{r} - 1}}{(1 - \hat{\theta}_{m}(\frac{1}{s}))^{q - 1}}\Biggl(\sum_{n = 1}^{\infty} \frac{1}{m + n}b_{n} \Biggr)^{q} \Biggr]^{\frac{1}{q}} > \frac{\pi}{\sin (\pi /r)}\Biggl(\sum _{n = 1}^{\infty} n^{\frac{q}{r} - 1} b_{n}^{q}\Biggr)^{\frac{1}{q}}, \end{aligned}
(31)

where $$\hat{\theta}_{m}(\frac{1}{s}): = \frac{\sin (\pi /s)}{\pi} \int _{0}^{\frac{1}{m}} \frac{u^{ - 1/r}}{1 + u}\,du = O(\frac{1}{m^{1/s}}) \in (0,1)$$ ($$m \in \mathrm{N}$$).

Inequality (29) is a reverse of (1).

4 Conclusions

In this paper, by virtue of the symmetry principle, applying the techniques of real analysis and Euler–Maclaurin summation formula, we construct proper weight coefficients and use them to establish a reverse extended Hardy–Hilbert’s inequality with multi-parameters in Lemma 3. Then, we obtain the equivalent forms and some equivalent statements of the best possible constant factor related to several parameters in Theorem 1 and Theorem 2. Finally, we illustrate how the obtained results can generate some new reverse Hardy–Hilbert-type inequalities in Remark 3. The lemmas and theorems provide an extensive account of this type of reverse inequalities.

Availability of data and materials

The data used to support the findings of this study are included within the article.

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Acknowledgements

The authors thank the referee for his useful proposal to reform the paper.

Funding

This work is supported by the National Natural Science Foundation (Nos. 11961021, 11561019), Guangxi Natural Science Foundation (2020GXNSFAA159084) and Hechi University Research Foundation for Advanced Talents under Grant (Nos. 2019GCC005, 2021GCC024). We are grateful for this help.

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BY carried out the mathematical studies, participated in the sequence alignment, and drafted the manuscript. RL and XH participated in the design of the study and performed the numerical analysis. All authors read and approved the final manuscript.

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Correspondence to Ricai Luo.

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Luo, R., Yang, B. & Huang, X. A reverse extended Hardy–Hilbert’s inequality with parameters. J Inequal Appl 2023, 58 (2023). https://doi.org/10.1186/s13660-023-02967-5